jmc

Multiplying both sides by $(x-5)(x-3{)}^2,$ we get $$4x=A(x-3{)}^2+B(x-5)(x-3)+C(x-5).$$Setting $x=5,$ we get $4A=20,$ so $A=5.$

Setting $x=3,$ we get $-2C=12,$ so $C=-6.$ Hence, $$4x=5(x-3{)}^2+B(x-5)(x-3)-6(x-5).$$Then $$B(x-5)(x-3)=-5x^2+40x-75=-5(x-3)(x-5),$$so $B=-5.$ Therefore, $(A,B,C)=\boxed{(5,-5,-6)}.$